Find 7'' (t), given 7'(t) = (te²t, − cos(5t + 5), 3t ln( − t)). ř'(t) = x, 4 cos(2t) = -6e-2t x, -3t² X

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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3.2.1

**Problem Statement:**

Find \(\vec{r}''(t)\), given \(\vec{r}(t) = \langle te^{2t}, -\cos(5t + 5), 3t \ln(-t) \rangle\).

**Solution Steps:**

1. **Determine the first derivative \(\vec{r}'(t)\):**  
   To find \(\vec{r}'(t)\), take the derivative of each component of \(\vec{r}(t)\).

2. **Calculate the second derivative \(\vec{r}''(t)\):**  
   Differentiate each component of \(\vec{r}'(t)\) to find \(\vec{r}''(t)\).

**Errors in Provided Solution:**

The proposed solution for \(\vec{r}''(t)\) was:

\[
\langle -6e^{-2t}, 4\cos(2t), -3t^2 \rangle \quad \text{(incorrect)}
\]

The incorrect components should be recalculated by taking the second derivatives of the original function's components:

- The first component of \(\vec{r}(t)\) is \(te^{2t}\).
- The second component of \(\vec{r}(t)\) is \(-\cos(5t + 5)\).
- The third component of \(\vec{r}(t)\) is \(3t \ln(-t)\).

**Note:**
Each of these calculations needs to be corrected to find the accurate \(\vec{r}''(t)\).

This task requires careful reevaluation of the differentiation process to correct these errors.
Transcribed Image Text:**Problem Statement:** Find \(\vec{r}''(t)\), given \(\vec{r}(t) = \langle te^{2t}, -\cos(5t + 5), 3t \ln(-t) \rangle\). **Solution Steps:** 1. **Determine the first derivative \(\vec{r}'(t)\):** To find \(\vec{r}'(t)\), take the derivative of each component of \(\vec{r}(t)\). 2. **Calculate the second derivative \(\vec{r}''(t)\):** Differentiate each component of \(\vec{r}'(t)\) to find \(\vec{r}''(t)\). **Errors in Provided Solution:** The proposed solution for \(\vec{r}''(t)\) was: \[ \langle -6e^{-2t}, 4\cos(2t), -3t^2 \rangle \quad \text{(incorrect)} \] The incorrect components should be recalculated by taking the second derivatives of the original function's components: - The first component of \(\vec{r}(t)\) is \(te^{2t}\). - The second component of \(\vec{r}(t)\) is \(-\cos(5t + 5)\). - The third component of \(\vec{r}(t)\) is \(3t \ln(-t)\). **Note:** Each of these calculations needs to be corrected to find the accurate \(\vec{r}''(t)\). This task requires careful reevaluation of the differentiation process to correct these errors.
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